Monrad.410.03_hw1solution

Monrad.410.03_hw1solution - STAT mo / MATH 494 EXuc/ise,...

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Unformatted text preview: STAT mo / MATH 494 EXuc/ise, 1.7.? a) : xg(44ix)g(%)$(g)(4_$) Since P(X g 0) = f(0) = j—§(§)0(§)4 z (3)4 = 0.3164063 < 0.5 and P(X g 1): f(0) +f(1) = 0.3164063+ fia(§)(§)3 : 0.3164063+0.421875 : = 0.7382813 > 0.5 the median is 1. b) 2 3:1:2 0 < ac < 1 The median m satisfies 0.5 = P(X S m) : fom : fom 3$2d$ : x3l6” : m3 => m z (0.5);7 c) = «(1:62), -oo < at < 00 Using the substitution 3/ 2 ~33, dy : —da: we get: P(X g 0) 2 f3“, 7r“Lavina : fooo Wdy 2 P(X > 0) Since P(X g 0) + P(X > O) = 1 both probabilities equai 0.5 = P(X g 0) = P(X > 0) proving 0 is the median. More general 0 is the median of any distribution with symmetric density function. xeAc/ise, 1.7.2H f(x)=§,—1g_:cg ,Y=X2. 3 Forl<y<4,G(y)=P(Y§y)=P(X2§y)=P—1£XS\/§)= =Mf(w)dx=ff%dw=§l E H Q + H ‘6 A (Q i V II Q A (I: V II In Exes/1013c [.01. 2. f(:c) = (am, a; =1,2,3... 00 3’2; 6‘ oo eta;__e’1 _et Ma) 2 EKG“) 2 :21 f($)etx = 12:1(3) = 723:0(3) — 71_ t — 2:39 for 532: < 1 that is t < 172(2). Then , 6:! _et) e2t e: __ _ 26‘! 2—6t [2+26t12-‘6t [2st = M”(O) : %(23:tt)2 it=O 2- (2_et)4 |t=0 : 1—4 2 6 mo = E0?) - [E0012 = 6 — 22 = 2 Exam/13¢ 127.2! TM wsmwm mew 0mg ' 1346 = Fae-t} =-§+x(«)a4x = 1—e“‘tj Mao, TM figs/figukm Malian 0? Y: X2 is Emma /. 916? W wMo/ls W3 [crab/W4 done Aka EKMMPQ /.9.L/.: V M(t) : (1 — t)‘3 Starting with the well known equality: _ = 2:20 ti and taking derivatives twice on both sides we get: (1—2t)3 : :2_1)t(i_2) = 223104“ + llti 01” Zlflt) :. (1'—— t)‘3 :: 3:0 LXUZZ'H it On the other hand M(t) : :00 4M“) 02“ I 229:0 Eli‘zt‘. i=0 1'1. , . Together these show 2 i! H2 * 9:223, i=1,2,3,.... A simPZe/IJ mm ng/éffinmd gala/ion (g; Mm %< 1 . N"(t5 f 3-H(r—t>’5 : £41, Way 1-. s.L/~.-(n+z>(r-£>'("*3) a 1541, EM) = We) =- Mwmz) mags)... Exam/Ge, L109 In Theorem 1.10.2 (Martha’s Inefiwc/i > wt (a/q uOO = etx and c = e“. This dam/gm a4 Pr[etX Z em] 3 13—53273 = e"t“M(t). For t > 0 the events 6“ 2 em and X Z a are equivalent hence P(X Z a) = Pr[etX Z em] 3 e—t“M(t) For t < 0 the events etX 2 em and X S a are equivalent hence P(X S a) : Pr[etX Z 6”] S e‘t“M(t) H (J~3 Q l v}. v .4: ...
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Monrad.410.03_hw1solution - STAT mo / MATH 494 EXuc/ise,...

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